Question
An observer, $1.5m$ tall, is $28.5m$ away from a $30m$ high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer


Let $AB = 1.5m$ be the observer and $CD = 30m$ be the tower.
Let the angle of elevation of the top of the tower be $\alpha.$
$CD = CE + ED$
$\Rightarrow CD = CE + AB$
$\Rightarrow 30 = CE + 1.5$
$\Rightarrow CE = 30 - 1.5 = 28.5m$
In $\triangle\text{CEB,}$
$\tan\alpha=\frac{\text{CE}}{\text{BE}}=\frac{28.5}{28.5}$
$\Rightarrow\ \tan\alpha=1$
$\Rightarrow\ \tan\alpha=\tan45^\circ$
$\Rightarrow\ \alpha=45^\circ$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free